IF ` f(x)= sin^2 x + cosec ^2 x , ` then the minimum value of f(x) is

Options

A. 1
B. `1.5`
C. `2`
D. 3

Answer

To find the minimum value of the function \( f(x) = \sin^2 x + \csc^2 x \), we can follow these steps:

### Step 1: Rewrite the function
We know that \( \csc x = \frac{1}{\sin x} \). Therefore, we can rewrite \( f(x) \) as:
\[ f(x) = \sin^2 x + \frac{1}{\sin^2 x} \]

### Step 2: Let \( y = \sin^2 x \)
Since \( \sin^2 x \) is always between 0 and 1 (inclusive), we can let \( y = \sin^2 x \). Thus, we can express \( f(x) \) in terms of \( y \):
\[ f(y) = y + \frac{1}{y} \]
where \( 0 < y \leq 1 \).

### Step 3: Find the derivative
To find the minimum value of \( f(y) \), we can take the derivative and set it to zero:
\[ f'(y) = 1 - \frac{1}{y^2} \]
Setting the derivative equal to zero gives:
\[ 1 - \frac{1}{y^2} = 0 \implies \frac{1}{y^2} = 1 \implies y^2 = 1 \implies y = 1 \]

### Step 4: Evaluate the second derivative
To confirm that this point is a minimum, we can check the second derivative:
\[ f''(y) = \frac{2}{y^3} \]
Since \( y > 0 \), \( f''(y) > 0 \), indicating that \( f(y) \) is concave up at \( y = 1 \).

### Step 5: Calculate the minimum value
Now we evaluate \( f(y) \) at \( y = 1 \):
\[ f(1) = 1 + \frac{1}{1} = 2 \]

### Conclusion
Thus, the minimum value of \( f(x) = \sin^2 x + \csc^2 x \) is:
\[ \boxed{2} \]